新課標(biāo)2022年版高考數(shù)學(xué)總復(fù)習(xí)第二章函數(shù)第六節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)練習(xí)含解析文

PAGEPAGE15第六節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)學(xué)習(xí)要求:1.理解對(duì)數(shù)的概念及其運(yùn)算性質(zhì),知道用換底公式能將一般對(duì)數(shù)轉(zhuǎn)化成自然對(duì)數(shù)或常用對(duì)數(shù);了解對(duì)數(shù)在化簡(jiǎn)運(yùn)算中的作用.2.理解對(duì)數(shù)函數(shù)的概念,理解對(duì)數(shù)函數(shù)的單調(diào)性,掌握對(duì)數(shù)函數(shù)圖象通過(guò)的特殊點(diǎn).3.知道對(duì)數(shù)函數(shù)是一類重要的函數(shù)模型.4.了解指數(shù)函數(shù)y=ax與對(duì)數(shù)函數(shù)y=logax互為反函數(shù)(a>0,且a≠1).1.對(duì)數(shù)的概念(1)對(duì)數(shù)的定義:一般地,如果①ax=N(a>0,且a≠1),那么數(shù)x叫做以a為底N的對(duì)數(shù),記作②x=logaN,其中③a叫做對(duì)數(shù)的底數(shù),④N叫做真數(shù).

(2)幾種常見的對(duì)數(shù):對(duì)數(shù)形式特點(diǎn)記法一般對(duì)數(shù)底數(shù)為a(a>0,且a≠1)⑤logaN

常用對(duì)數(shù)底數(shù)為10⑥lgN

自然對(duì)數(shù)底數(shù)為e⑦lnN

2.對(duì)數(shù)的性質(zhì)與運(yùn)算法則(1)對(duì)數(shù)的性質(zhì):alogaN=⑧N;logaaN=⑨N.(a>0,且(2)對(duì)數(shù)的重要公式:換底公式:⑩logbN=logaNlogab(a相關(guān)結(jié)論:logab=1logba,logab·logbc·logcd=logad(a,b,c均大于0且不等于1,d(3)對(duì)數(shù)的運(yùn)算法則:如果a>0且a≠1,M>0,N>0,那么logaMN=logaM+logaN;

logaMN=logaM-logaN;

logaMn=nlogaM(n∈R);

logamMn=nmlogaM(m,n3.對(duì)數(shù)函數(shù)的圖象與性質(zhì)a>10<a<1圖象性質(zhì)定義域:(0,+∞)值域:R圖象恒過(guò)點(diǎn)(1,0),即x=1時(shí),y=0當(dāng)x>1時(shí),y>0;當(dāng)0<x<1時(shí),y<0當(dāng)x>1時(shí),y<0;當(dāng)0<x<1時(shí),y>0是(0,+∞)上的增函數(shù)是(0,+∞)上的減函數(shù)4.反函數(shù)指數(shù)函數(shù)y=ax(a>0,且a≠1)與對(duì)數(shù)函數(shù)y=logax(a>0,且a≠1)互為反函數(shù),它們的圖象關(guān)于直線y=x對(duì)稱.

知識(shí)拓展對(duì)數(shù)函數(shù)的圖象與底數(shù)大小的比較如圖,作直線y=1,則該直線與四個(gè)函數(shù)圖象交點(diǎn)的橫坐標(biāo)為相應(yīng)的底數(shù),故0<c<d<1<a<b.由此我們可得到以下規(guī)律:在第一象限內(nèi),從左到右底數(shù)逐漸增大.1.判斷正誤(正確的打“√”,錯(cuò)誤的打“?”).(1)logaMN=logaM+logaN. ()(2)logax·logay=loga(x+y). ()(3)log2x2=2log2x. ()(4)若logam<logan,則m<n. ()(5)函數(shù)y=ln1+x1-x與y=ln(1+x)-ln(1-x)的定義域相同(6)對(duì)數(shù)函數(shù)y=logax(a>0,且a≠1)的圖象過(guò)定點(diǎn)(1,0),且過(guò)點(diǎn)(a,1),1a,-1,函數(shù)圖象經(jīng)過(guò)第一、四象限.答案(1)?(2)?(3)?(4)?(5)√(6)√2.log525+1612= (A.94答案B3.下列各式中正確的是 ()A.loga6loga3=loga2C.(lnx)2=2lnxD.lg5x3=答案D4.已知a=log23,b=1212,c=1313,則a,bA.a<b<cB.a<c<bC.b<c<aD.c<b<a答案D5.(2020河北唐山期末)函數(shù)f(x)=lg(x-2)的定義域?yàn)?()A.(-∞,+∞)B.(-2,2)C.[2,+∞)D.(2,+∞)答案D6.(易錯(cuò)題)已知a>0,且a≠1,則函數(shù)f(x)=ax與函數(shù)g(x)=logax的圖象可能是 ()答案B由函數(shù)f(x)=ax與函數(shù)g(x)=logax互為反函數(shù),得圖象關(guān)于y=x對(duì)稱,從而排除A,C,D.易知當(dāng)a>1時(shí),兩函數(shù)圖象與B選項(xiàng)中的圖象相同.故選B.易錯(cuò)分析忽視反函數(shù)的定義.對(duì)數(shù)的概念、性質(zhì)與運(yùn)算角度一對(duì)數(shù)的概念與性質(zhì)典例1(1)若loga2=m,loga5=n(a>0,且a≠1),則a3m+n= ()A.11B.13C.30D.40(2)已知2a=5b=10,則a+bab(3)設(shè)52log5(2x答案(1)D(2)1(3)2角度二對(duì)數(shù)的運(yùn)算典例2計(jì)算:(1)(lg2)2+lg2·lg50+lg25;(2)log34273+lg5+7log72+log23·(3)(log32+log92)·(log43+log83).解析(1)原式=(lg2)2+(1+lg5)·lg2+lg52=(lg2+lg5+1)·lg2+2lg5=(1+1)·lg2+2lg5=2(lg2+lg5)=2.(2)原式=log3334-1+lg5+2+=-14(3)原式=log32·log43+log32·log83+log92·log43+log92·log83=lg2lg3·lg32lg2+lg2lg3·lg3規(guī)律總結(jié)對(duì)數(shù)運(yùn)算的求解思路(1)首先利用冪的運(yùn)算把底數(shù)或真數(shù)進(jìn)行變形,化成分?jǐn)?shù)指數(shù)冪的形式,使冪的底數(shù)最簡(jiǎn),然后利用對(duì)數(shù)的運(yùn)算性質(zhì)求解.(2)將對(duì)數(shù)式化為同底數(shù)對(duì)數(shù)的和、差、倍數(shù)運(yùn)算,然后逆用對(duì)數(shù)的運(yùn)算性質(zhì),將其轉(zhuǎn)化為同底數(shù)對(duì)數(shù)真數(shù)的積、商、冪的運(yùn)算.1.(lg5)2+lg2·lg5+lg20-log23·log38+2(1+log答案9解析原式=lg5·(lg5+lg2)+lg2+lg10-log23·log28log23+22.如果45x=3,45y=5,那么2x+y=.

答案1解析∵45x=3,45y=5,∴x=log453,y=log455,∴2x+y=2log453+log455=log459+log455=log45(9×5)=1.對(duì)數(shù)函數(shù)的圖象及應(yīng)用典例3(1)函數(shù)f(x)=ln|x-1|的大致圖象是 ()(2)當(dāng)0<x≤12時(shí),4x<logax(a>0,且a≠1),則a的取值范圍是 (A.0C.(1,2)(3)已知函數(shù)f(x)=4+loga(x-1)(a>0,且a≠1)的圖象恒過(guò)定點(diǎn)P,則點(diǎn)P的坐標(biāo)是.

答案(1)B(2)B(3)(2,4)方法技巧對(duì)數(shù)函數(shù)圖象的應(yīng)用方法一些對(duì)數(shù)型方程、不等式的問(wèn)題常轉(zhuǎn)化為相應(yīng)函數(shù)的圖象問(wèn)題,利用數(shù)形結(jié)合求解.1.(2020黑龍江齊齊哈爾第六中學(xué)模擬)函數(shù)f(x)=|loga(x+1)|(a>0,且a≠1)的大致圖象是 ()答案C函數(shù)f(x)=|loga(x+1)|的定義域?yàn)閧x|x>-1},且對(duì)任意的x∈(-1,+∞),均有f(x)≥0,結(jié)合對(duì)數(shù)函數(shù)的圖象可知選C.2.函數(shù)y=x-a與函數(shù)y=logax(a>0,且a≠1)在同一坐標(biāo)系中的圖象可能是 ()答案C當(dāng)a>1時(shí),對(duì)數(shù)函數(shù)y=logax為增函數(shù),當(dāng)x=1時(shí),函數(shù)y=x-a的值為負(fù),故A、D錯(cuò)誤;當(dāng)0<a<1時(shí),對(duì)數(shù)函數(shù)y=logax為減函數(shù),當(dāng)x=1時(shí),函數(shù)y=x-a的值為正,故B錯(cuò)誤,C正確.故選C.對(duì)數(shù)函數(shù)的性質(zhì)及應(yīng)用角度一比較對(duì)數(shù)值的大小典例4(1)(2018天津,5,5分)已知a=log2e,b=ln2,c=log1213,則a,b,cA.a>b>cB.b>a>cC.c>b>aD.c>a>b(2)已知f(x)滿足f(x)-f(-x)=0,且在(0,+∞)上單調(diào)遞減,若a=79-14,b=9715,c=log219,則f(a),A.f(b)<f(a)<f(c)B.f(c)<f(b)<f(a)C.f(c)<f(a)<f(b)D.f(b)<f(c)<f(a)答案(1)D(2)C角度二解簡(jiǎn)單的對(duì)數(shù)不等式典例5(1)函數(shù)f(x)=1(log2A.0,1C.0,12∪(2,+∞)D.(2)函數(shù)y=log3(2A.[1,2]B.[1,2)C.2答案(1)C(2)C角度三對(duì)數(shù)函數(shù)性質(zhì)的綜合應(yīng)用典例6已知函數(shù)f(x)=loga(ax2-x+1)(a>0,且a≠1).(1)若a=12,求函數(shù)f(x)的值域(2)當(dāng)f(x)在區(qū)間14,32上為增函數(shù)時(shí),解析(1)當(dāng)a=12時(shí),ax2-x+1=故函數(shù)f(x)的定義域?yàn)镽,∵12x2-x+1=12[(x-1)2+1]≥12,∴l(xiāng)og1212即函數(shù)f(x)的值域?yàn)?-∞,1].(2)依題意可知,①當(dāng)a>1時(shí),由復(fù)合函數(shù)的單調(diào)性可知,必有y=ax2-x+1在14,32上單調(diào)遞增,且ax2-x+1>0對(duì)任意的x所以x=12a②當(dāng)0<a<1時(shí),同理可得必有y=ax2-x+1在14,32上單調(diào)遞減,且ax2-x+1>0對(duì)任意的x所以x=12a≥綜上,a的取值范圍是29,1規(guī)律總結(jié)1.比較對(duì)數(shù)值大小的方法(1)若底數(shù)為同一常數(shù),則可由對(duì)數(shù)函數(shù)的單調(diào)性直接進(jìn)行判斷;若底數(shù)為同一字母,則需對(duì)底數(shù)進(jìn)行分類討論.(2)若底數(shù)不同,真數(shù)相同,則可以先用換底公式化為同底后,再進(jìn)行比較.(3)若底數(shù)與真數(shù)都不同,則常借助1,0等中間量進(jìn)行比較.2.對(duì)數(shù)不等式的類型及解法(1)形如logax>logab(a>0,且a≠1)的不等式,需借助y=logax的單調(diào)性求解,如果a的取值不確定,那么需分a>1與0<a<1兩種情況討論.(2)形如logax>b(a>0,且a≠1)的不等式,需先將b化為以a為底的對(duì)數(shù)式的形式,再求解.1.設(shè)a=log36,b=log510,c=log714,則 ()A.c>b>aB.b>c>aC.a>c>bD.a>b>c答案D∵a=log36=1+log32=1+1log23,b=log510=1+log52=1+1log22.(2019山東高考模擬)已知f(x)=ex-1+4x-4,若正實(shí)數(shù)a滿足floga34<1,則aA.a>3C.0<a<34或a>1D.a答案C因?yàn)閥=ex-1與y=4x-4都是R上的增函數(shù),所以f(x)=ex-1+4x-4是R上的增函數(shù),又因?yàn)閒(1)=e1-1+4-4=1,所以flog所以loga34<logaa當(dāng)0<a<1時(shí),y=logax在(0,+∞)上單調(diào)遞減,所以a<34當(dāng)a>1時(shí),y=logax在(0,+∞)上單調(diào)遞增,所以a>34,故a綜上所述,a的取值范圍是0<a<34或a>1.故選3.(2020上海高三專題練習(xí))函數(shù)y=log0.答案-14,解析由題意可知0<4x2-3x≤1,解得x∈-14,4.(2020河北新樂(lè)一中高二月考)函數(shù)f(x)=log13(-x2+2x+3)的單調(diào)遞增區(qū)間是答案[1,3)解析令u=-x2+2x+3,由u>0,解得-1<x<3,即函數(shù)f(x)的定義域?yàn)?-1,3),根據(jù)二次函數(shù)的圖象與性質(zhì)可知函數(shù)u=-x2+2x+3在(-1,1]上單調(diào)遞增,在[1,3)上單調(diào)遞減,因?yàn)楹瘮?shù)f(x)=log13u所以根據(jù)復(fù)合函數(shù)同增異減可得函數(shù)f(x)的單調(diào)遞增區(qū)間為[1,3).5.已知函數(shù)f(x)=ln(1+9x2解析由1+9x2-3x>0恒成立知函數(shù)f(x)的定義域?yàn)橐驗(yàn)閒(-x)+f(x)=[ln(1+9x2+3x)+1]+[ln(1+9x2-3x)+1]=ln所以f(lg2)+flg12=f(lg2)+f(-lgA組基礎(chǔ)達(dá)標(biāo)1.已知函數(shù)f(x)=log2(x2-2x+a)的最小值為2,則a= ()A.4B.5C.6D.7答案B2.log29×log34+2log510+log50.25= ()A.0B.2C.4D.6答案D3.(2020河北冀州中學(xué)模擬)函數(shù)y=log3(2A.[1,2]B.[1,2)C.2答案C4.log6[log4(log381)]的值為 ()A.-1B.1C.0D.2答案C5.(2019河南鄭州模擬)設(shè)a=log50.5,b=log20.3,c=log0.32,則 ()A.b<a<cB.b<c<aC.c<b<aD.a<b<c答案B6.若lg2=a,lg3=b,則log418= ()A.aC.a答案D7.已知函數(shù)f(x)=lg1-x1+x,若f(aA.2B.-2C.1答案D8.設(shè)f(x)=lg(10x+1)+ax是偶函數(shù),則a的值為 ()A.1B.-1C.1答案D函數(shù)f(x)=lg(10x+1)+ax的定義域?yàn)镽,因?yàn)閒(x)為偶函數(shù),所以f(x)-f(-x)=0,即lg(10x+1)+ax-[lg(10-x+1)+a(-x)]=(2a+1)x=0,所以2a+1=0,解得a=-12.B組能力拔高9.已知f(x)=log12x,則不等式(f(x))2>f(x2)的解集為 (A.0,1C.14,答案D由(f(x))2>f(x2)得(log12x)2>log12x2?log12x(log12x-2)>0,即log110.若x、y、z均為正數(shù),且2x=3y=5z,則 ()A.2x<3y<5zB.5z<2x<3yC.3y<5z<2xD.3y<2x<5z答案D令2x=3y=5z=k(k>1),則x=log2k,y=log3k,z=log5k,∴2x3y=2lgklg2·lg33lg2x5z=2lgklg2·lg55lgk=11.(2020福建莆田模擬)已知函數(shù)f(x)=|log3x|,實(shí)數(shù)m,n滿足0<m<n,且f(m)=f(n),若f(x)在[m2,n]上的最大值為2,則nm=答案9解析∵f(x)=|log3x|,實(shí)數(shù)m,n滿足0<m<n,且f(m)=f(n),∴0<m<1<n,-log3m=log3n,∴mn=1.∵f(x)在區(qū)間[m2,n]上的最大值為2,且函數(shù)f(x)在[m2,1)上是減函數(shù),在(1,n]上是增函數(shù),∴-log3m2=2或log3n=2.若-log3m2=2,則m=13(舍負(fù)),故n此時(shí)log3n=1=-log3m,符合題意,即nm若log3n=2,則n=9,故m=19,此時(shí)-log3m2=4>2,不符合題意.故nmC組思維拓展12.(2020四川攀枝花模擬)設(shè)函數(shù)f(x)=|logax|(0<a<1)的定義域?yàn)閇m,n](m<n),值域?yàn)閇0,1],若n-m的最小值為13,則實(shí)數(shù)a的值為答案23解析作出y=|logax|(0<a<1)的大致圖象如圖所示,令|logax|=1,得x=a或x=1a,13.若loga(a2+1)<loga(2a)<0